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Be atom splitting into two helium atoms!

  1. Jan 22, 2013 #1
    Simple and stupid question!
    A Be atom is traveling with 60 kev kinetic energy and splits into two helium atoms, and the process itself releases 92.2 kev. One helium atom moves at a 30 degree angle with respect to x. find the direction of motion of the second helium atom and find the velocity of each helium atom.
    My problem: to get the y component info, i feel like i have to assume the other helium atom is traveling at -30 degree angle and that both he atoms have the same velocity. is this wrong? if so, how would i go about solving this?
    So far i have 7.338 x 10^12 = V1^2 + V2^2 and sin(30)V1 = -sin(theta)v2
    Thank yoU!
     
  2. jcsd
  3. Jan 22, 2013 #2

    Simon Bridge

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    ... where does this energy go?

    Anyway, you are using conservation of (total!) energy and momentum ... you shouldnt have to assume a direction, it should be dictated by the physics.

    Presumably you found:
    http://answers.yahoo.com/question/index?qid=20090126134949AAXPK4B
    ... I'd put energy in keV or MeV and masses in keV/c2 or MeV/c2 instead of messing about with SI units.
     
  4. Jan 22, 2013 #3

    tms

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    Where did the 7.338 x 10^12 come from? What physical principles are you using?
     
  5. Jan 22, 2013 #4
    I converted everything from keV to J. So the total energy in the system is 60 keV + 92.2 keV, which equals .5mv1^2 + .5mv2^2. Converting the keV to J, and rewriting it (m = 4u = 6.645 x 10^-27 kg), you get 7.338 x 10^12 = v1^2 + v2^2. And the professor wants us to use SI right away so thats why I converted.
    I also used conservation of momentum to get:
    m(sin30)v1 + m(sin(theta))v2 = 0 (because initial y momentum is 0)
    and
    m(cos(30))v1 + m(cos(theta))v2 = initial x momentum = 1.6 x 10^-20.
    My problem is I cant figure out how to solve for either theta or v1 or v2. I just need one of these values to get the rest.
     
  6. Jan 22, 2013 #5

    haruspex

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    You have 3 equations and 3 unknowns, so you've a reasonable hope of a solution.
    Use, say, the y-momentum eqn to substitute for v1 in each of the other two, and so on. (But there might be quicker ways.)
     
  7. Jan 23, 2013 #6
    Couldnt you just say by symmetry the other He atom must be at -30 degrees with the same velocity?
     
  8. Jan 23, 2013 #7

    tms

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    Look at things in the CM frame.
     
  9. Jan 23, 2013 #8

    haruspex

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    I don't see any basis for saying it's symmetric. Think of the Be atom as two balls, with the line joining their centres not aligned with the direction of movement. If they blow apart, the movements will not be symmetric about the original line of travel.
    Please, just try the obvious and standard process of successive substitution to solve the simultaneous equations, or see if recasting them in the CoM frame as tms suggests simplifies them.
     
  10. Jan 23, 2013 #9

    Simon Bridge

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    I think that question is a sign you'd better just do the math like the others suggest. Why so reluctant with the simultaneous equations?

    note: sin(30)=1/2, cos(30)=√3/2
     
  11. Jan 27, 2013 #10
    I am reluctant with the simultaneous equations because I can not solve them. I did substitution and can not break it down far enough in order to get an answer. Also, the second part of the question is using the CM frame so that can not be used in this instance. Please help my friend and I are banging our heads together on this one
     
  12. Jan 27, 2013 #11

    tms

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    Show your work.
     
  13. Jan 27, 2013 #12

    Simon Bridge

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    Then you need the practice ;) please show your working then we can unstick you.
    If you don't show us your working so far, we cannot help you.
     
  14. Jan 27, 2013 #13

    haruspex

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    You really should at least post your attempt, but I think I see why it gets tricky.
    You have
    1) v12+v22=E (known constant)
    2) sin(30)v1 + sin(θ)v2 = 0
    3) cos(30)v1 + cos(θ)v2 = P (known constant)
    Substituting for v2 based on (2):
    1') v12+(sin(30)v1cosec(θ))2 = E = v12(1+cosec2(θ)/4)
    3') cos(30)v1 - cos(θ)v1sin(30)cosec(θ) = P = v1(cos(30)-cot(θ)/2)
    E/(P2) = an expression in which the only unknown is θ.
    This can be written in the form sin2(α)+sin2(θ) = k sin2(θ-α), where α=30 degrees, or converted to a quartic in sin(θ). I don't see any neat way to solve it. Simplest may be to plot a curve against θ and see where the function takes the right value, or perhaps you have software that will solve it.
     
  15. Jan 28, 2013 #14
    I believe i figured it out.
    Original y momentum is 0, msin(30)v1 + msin(theta)v2 = 0. therefore: -v1/2 = Vy2.
    Then 1.6 x 10^-20 = m(cos(30)V1 + cos(theta)V2)
    then 2.4 x 10^6 = v1(√3)/2 + cos(theta)V2
    then: Vx2 = 2.4E6 - V1√3/2
    Then V2 = √(Vy2^2 + Vx2^2)
    Which can then be used with the energy equation: 7.338E12 = V1^2 + V2^2.
    Everything is in terms of V1, solve and get two solutions. Calculate each V1s V2 counterpart, plug in both to 7.338E12 = V1^2 + V2^2. and only one checks out.
    V1= -320189
    V2 = 2689885.6
    Theta is tan^-1(V2y/V2x) = 3.412 degrees.

    I checked the numbers and they all seem to work.
    Moving on, the next part of the question somewhat confuses me.
    Consider one He atom to be emitted with veloicty components of Vx and Vy in the Be rest frame. What is the relationship between Vx and Vy (the angle right?!?) How do vx and vy change when we move in the x direction at speed v? I think it wants me to use the frame of reference with a speed of the original Be atom before it splits. This kinda confuses me though. Any insight would be appreciated.
     
  16. Jan 28, 2013 #15

    tms

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    Where does that number come from?
    Where does that number come from?
    Where does that number come from?

    When working problems, it is always much better to solve them symbolically, and only plug in numbers at the very end. Plugging in numbers early throws away information, such as where the resulting number came from. It also makes it harder for anyone to follow your reasoning; instead of following along with your algebra, your readers must guess at what you might have done. If you are correct, the guessing is easier, but if you have made a mistake, the reader has to guess what mistake you made.
     
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